Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x - 4} = \dfrac{16}{x - 4}$
Solution: Multiply both sides by $x - 4$ $ \dfrac{x^2}{x - 4} (x - 4) = \dfrac{16}{x - 4} (x - 4)$ $ x^2 = 16$ Subtract $16$ from both sides: $ x^2 - (16) = 16 - (16)$ $ x^2 - 16 = 0$ Factor the expression: $ (x - 4)(x + 4) = 0$ Therefore $x = 4$ or $x = -4$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.